Q.If $x=f(t)$ and $y=g(t)$ are differentiable functions of $t$ so that $y$ is a function of $x$ and if $\dfrac{dx}{dt}\ne 0$ then prove that $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$. Hence find the derivative of $7^x$ w.r.t. $x^7$.

Prove the parametric-differentiation chain rule, then apply it with $u=7^x$ and $v=x^7$ as functions of $t=x$.

Proof. Let $\delta x$ be a small increment in $x$, causing increments $\delta y$ in $y$ and $\delta t$ in $t$ correspondingly (since $x=f(t)$, $y=g(t)$). As $\delta t \to 0$, both $\delta x\to0$ and $\delta y\to0$ (by continuity/differentiability of $f,g$).

$\frac{\delta y}{\delta x}=\frac{\delta y/\delta t}{\delta x/\delta t}$

Taking $\delta t\to0$ (hence $\delta x\to0$, since $dx/dt\ne0$ means $x$ is locally invertible / $\delta x\ne 0$ for small $\delta t$):

$\frac{dy}{dx}=\lim_{\delta t\to0}\frac{\delta y/\delta t}{\delta x/\delta t}=\frac{dy/dt}{dx/dt} \qquad (\text{since } dx/dt e0)$

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